text:
What effect does a direct crosswind have on takeoff performance?
Is the take-off distance increased by a direct crosswind?

Are you asking about a 90 degree crosswind specifically, or any crosswind?

Yes, 90 degree crosswind.

Any crosswind is the sum of a sideways, 90º component and a component parallel to the runway. In other words, any crosswind has a 90º component...

@xxavier Yes, but it's possible, in a direct crosswind, for the parallel component to be zero. That's what the OP is asking about.

Consider - as a cyclist, riding across a 90 degree side wind is not the same as riding in a dead-calm.

@Criggie -- consider also -- if in a sailboat, you'd rather have a 90-degree crosswind than no wind -- :)

It is important to analyze the aerodynamics involved to understand what forces impact the aircraft during takeoff, rather than to rely on  flight operations manuals for specific aircraft to generalize what the effects might be on all classes of aircraft.

The effects of a constant crosswind up to the demonstrated value are insignificant (at least for airliners), as ymb1's answer already says. I checked some Boeing FCOMs and crosswind is never mentioned for takeoff performance calculation. It can therefore not have a significant impact. Boeing recommends this technique:

Any deviation from the centerline
during thrust application should be countered with immediate smooth and positive
control inputs. Smooth rudder control inputs combined with small control wheel
inputs result in a normal takeoff with no overcontrolling. [...]
Note: Excessive control wheel displacement during rotation and liftoff increases
spoiler deployment. As spoiler deployment increases, drag increases and
lift is reduced which results in reduced tail clearance, a longer takeoff roll,
and slower airplane acceleration.
[emphasis mine]

A gusty crosswind may however require larger control wheel inputs, which result in spoiler deployment. This will impact takeoff performance. Boeing therefore recommends to use more than the minimum required thrust in these situations:

For takeoff in gusty or strong crosswind conditions, use of a higher thrust setting
than the minimum required is recommended. When the prevailing wind is at or
near 90° to the runway, the possibility of wind shifts resulting in gusty tailwind
components during rotation or liftoff increases. During this condition, consider the
use of thrust settings close to or at maximum takeoff thrust.

Source: Boeing 737 NG FCTM 3.15 - Takeoff and Initial Climb - Crosswind Takeoff

To analyze the impact of a crosswind takeoff on performance, it is necessary to break down the effects on each phase of the maneuver.  For simplicity, let's examine a single engine trainer type aircraft, and see where the crosswind impacts takeoff performance.
At the start of the takeoff roll, with a crosswind, full deflection of aileron is made into the wind. This presents increased drag because both the left and right aileron are in a high deflection position.  Of course, at the initial takeoff roll, the drag is minimized because of the low airspeed.  As air speed increases, the deflection is reduced, which will have a limiting effect on the aileron drag, but that drag is still present, measurable and duplicatable.  Additionally, while rudder is used to maintain directional control, and counteract asymmetrical propeller trust, it is also being used to effectively maintain an aerodynamic slip while on the ground. Slips create increased drag, and slightly more power is consumed by the slip, being a combination of aileron input, rudder input and side forces on the fuselage.
After the plane departs the runway, the slip can transition to a crab, and the power losses associated with a slip disappear, as the aircraft is now in coordinated flight.
The impact on a crosswind takeoff, assuming the above technique, is on the takeoff roll, where the aircraft is in a slip.  The longer takeoff roll will increase the amount of runway used for the roll.  However, the climb out performance will not be affected adversely by the crosswind.  In fact the aggregate wind vector on the aircraft will be influenced by the crosswind, and may result in a higher "head wind" vector which would reduce the distance to clear an obstacle, or to get to altitude.
A simple summary would be the crosswind takeoff adversely impacts the runway roll, and the initial climb to clear a near obstacle such as power lines at the end of the runway.  However, the distance to climb to altitude may be reduced by some small amount. Climb rates are affected by power and once the power lost by the slip during the ground roll is regained, the aircraft will perform normally.
Both of these effects are readily model-able. Effects will of course be greater on slower aircraft, such as motogliders and trainers.  The effects will be lost in the noise (no pun intended) on high performance aircraft such as fighter jets.


text:
Relationship between Load-Factor AND Gliding Ratio
I am wondering about the g-load and its relation to the so called gliding ratio.
I read this really good contribution here on this forum:
Does lift equal weight in a climb?
So the g load factor is defined as: g=Lift / Weight   and the gliding ratio= L/D

In a descend flight: what does a g-load factor of 0.5 have for a meaning, when there is no thrust force.

L=Lift ; D=Drag ; W=Weight ; gamma=gliding angle
When gliding, the vertical force components (_v) have to balance the Weight, so:
L_v + D_v = W
with
L_v=Lcos(gamma)
and
D_v=Dsin(gamma)
So:
Lcos(gamma)+Dsin(gamma) = G (1)
AND
g=L / G = 0.5  (2)
Insert (2) in(1) it comes to:
L/D=sin(gamma)/(2-cos(gamma))
a) Is this correct ?
b) When I want to have a gamma=10° this would mean that L/D=0.17  and for gamma=40° L/D=0.52087.
But on the other hand gamma has a relation ship L/D=1/tan(gamma) . So for gamma=40° it comes to L/D=1.19
I am really confused about that. Can someone help me?
Thank you all experts!
Lucas

If you are at .5 G you are not gliding, you are accelerating downwards quite rapidly!

Lucas, you mean a steady-state glide, right?  Not just that the prop stopped turning?

I think W=G? Seems you switched symbols halfway

I’m not totally clear on what you are trying to accomplish here, but to expand on my earlier comment:  If you are expecting some sort of steady state L/D glide output using .5 G you won’t get it.  .5 G is not steady state, nor is it gliding.  You are accelerating rapidly downwards.  Falling in other words, at an ever increasing speed at just half the rate of free fall.  Could this be the problem with your math?

I am wondering about the g-load and its relation to the so called gliding ratio.
...

So the g load factor is defined as: g=Lift / Weight

We'll adopt that definition for the purposes of this answer.  Note that that is not the entire apparent g-load felt by the aircraft and pilot.  It is just the component that is acting in what we might loosely call the "upward" direction in the aircraft's reference frame.  More precisely, it is  the component acting perpendicular to the flight path.  Which is approximately the same as the component that would be registered by a conventional (mechanical) panel-mounted g-meter.  No need to get too caught up in semantics here, as long as we clearly define what we are talking about.
(Another minor point would be to suggest that you settle on "g-load" or "load factor", rather than combining the terms into "g-load factor".  Even though the two terms arguably have the same meaning.)

and the gliding ratio= L/D

Yes, that's certainly true

In a descend flight: what does a g-load factor of 0.5 have for a meaning

If L/W is 0.5, the glide angle is (arccos) 0.5, or 60 degrees below horizontal.  For more, see the related ASE answer What produces thrust along the line of flight in a glider?

L=Lift ; D=Drag ; W=Weight ; gamma=gliding angle

When gliding, the vertical force components (_v) have to balance the Weight, so:
L_v + D_v = W
with
L_v=Lcos(gamma)
and
D_v=Dsin(gamma)
So:
Lcos(gamma)+Dsin(gamma) = G (1)

All true except the last-- it should be W, not G.

AND

g=L / G = 0.5  (2)

Same-- change "G" to "W" and it's correct

Insert (2) in(1) it comes to:

L/D=sin(gamma)/(2-cos(gamma))

a) Is this correct ?

I don't understand how you did this substitution.  Can you edit the question to explain better?  Anyway, it is not correct.  In our example above where gamma is 60 degrees, when I plug that into your equation, I get an L/D of 0.57778, when it should be 0.5.

b) When I want to have a gamma=10° this would mean that L/D=0.17

From What produces thrust along the line of flight in a glider? we can see that D/L = tan (gamma), where gamma is the glide angle, so L/D = 1 / tan (gamma). If gamma is 10 degrees, L/D is 1 / tan (gamma) = 1 /.1763 = 5.672.    The inverse of what you said.

and for gamma=40° L/D=0.52087.

No, the tangent of 40 degrees is 0.8391, and 1/.8391 = 1.1918.  So that's the L/D associated with a 40-degree glide angle.

But on the other hand gamma has a relation ship L/D=1/tan(gamma) . So for gamma=40° it comes to L/D=1.19

Yes

I am really confused about that. Can someone help me?

I'm not sure I follow exactly where the confusion arises, but I hope that helps.

Some recent corrections made to math.

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text:
Flying upside down without feeling it
A friend of mine told me that several years ago she flew on a two-seats general aviation aircraft. She does not remember the exact model, but she remembers well that the pilot flew upside down for a while and she did not feel anything different from the normal upright position.
She also noticed that a water in a bottle was stick to the bottom of the bottle.
I guess it means the pilot managed to provide a net 1 g of acceleration upwards, so the body and any other object experienced the usual head-to-toe force.
But to achieve that acceleration it should flew with 2 g of acceleration (to compensate the gravity first). She said the flight path was pretty straight.
Questions:

is my guess correct?
I'm sorry to cannot provide a specific model, but given the short description of the aircraft, what flight path should he follow to get that acceleration?

Likely answered in Can one fly up side down while a glass of water keeps full due to g-forces?. Also Can spatial disorientation in low visibility situations lead a pilot to fly upside down?

@mins as far as I understand those answers are about a loop. She said "the path was pretty straight". She also said it was a general aviation aircraft, not a military jet. Hence I assumed the speed cannot be so fast

You cannot fly in straight line and keep water in the glass. Some acceleration is required, and is provided by the turn. You missed that in the linked question: "For an aircraft at 160 km/h the radius of the turn must be smaller than 200 m. On the other hand for a fighter at 500 km/h, the radius may be as large as 2 km". At 160 km/h, on a circle of 200 m, a full turn is about 30 s (see table).

I didn't miss that. For this reason I said "it was a general aviation aircraft". I assumed a speed of about 300 km/h. Hence if the radius was about 1 km it takes just 30-40 seconds to complete the loop. So it should be clearly visible to the passenger.

Sure it is visible on daytime below the clouds.

The term "for a while" would normally infer at least several minutes, but that's highly unlikely, so basically unless she gave a time estimate like 15 seconds or whatever, we have no idea how long it lasted. Also with the distraction of flying inverted, presumably a new experience for her, the plane could easily be making changes in direction without her noticing. Also she didn't necessarily feel 1 g, water requires very little g-force to remain in the bottle, the same with her, she wasn't being pulled out of the seat so it felt normal, she may not have noticed a reduced g load in the seat.

@StevePemberton good points.

Your friend may be interested in this video featuring the amazing Bob Hoover doing a barrel roll while pouring water into a glass: https://www.youtube.com/watch?v=V9pvG_ZSnCc

unreal video @WayneConrad thanks!

This sounds like a barrel roll.
This is a corkscrew-like flight path in which the airplane rotates around an axis between its longitudinal and pitch axes by flying along a path which lies on the surface of a horizontal cylinder. Like this (source):

By not following a straight flight path, the airplane experiences a centrifugal force which counteracts gravity when flying inverted. A barrel roll can be interpreted as a combination of a regular roll an a loop. Like in a loop, the centrifugal force can be used to experience positive acceleration throughout the maneuver.
Since the airplane never experiences negative g loads, a barrel roll can be flown with almost all types of airplanes, even airliners. If you know whether the seating was side-by-side or in a tandem arrangement, we can try to narrow down the range of types.

For a generic GA aircraft, how much long (in time) should be the barrel roll in order to provide the 1 g acceleration?

I can try to ask her about the seats arrangement.

@Mark: Several seconds, details depend on flight speed. Yes, please ask her!

@MichaelHall It's not necessarily 45°, but you are right, it is not exactly the longitudinal axis. But mostly.

@MichaelHall The interpretation of a blend between roll and loop is spot on! I will update the answer.

I'm having difficulty connecting to http://www.aircraftrecognition.co.uk

If you use the 45° angle from the 1954 US Navy training film then you would be flying perpendicular to your original heading when fully inverted. A smaller angle would give a smaller change of heading, but I suppose it is still the same kind of maneuver. The conceptual difference between "mostly" in the direction of the axis and exactly in the direction of the axis seems very large to me, however, and I felt I had overcome a fundamental misunderstanding when I realized the reference line and cylinder axis were not parallel.

@DavidK, Thanks for the link!  That's a cool old film, from right around when my dad went through training.  I've since deleted my earlier comments, but yes... the target was 90 degrees off entry heading inverted at the top of the maneuver, making the axis 45 degrees off.  This is how the Navy taught me, and I presume it is the same still.  Of course there are an infinite number of iterations of a lazy roll between this and a pure aileron roll.

In case anyone has trouble visualizing, lay a strip of wrapping ribbon on a table.  Place a cardboard paper towel tube on top at a 45 degree angle across the ribbon.  (any cylinder will work) Then wrap the ribbon up over the top, and underneath again so it comes out going in the original direction.  At the top the ribbon should be 90 degrees off.  That's what a proper USN Barrel Roll should look like!  (in case you care...)

@DavidK The page from where I copied the picture (http://ffden-2.phys.uaf.edu/webproj/211_fall_2016/Collin_Lasley/collin_lasley/AileronRolls.html) referenced it to aircraftrecognition.co.uk. I never checked the link, but yes, I have trouble, too.

Remark: if you truly follow a path on a cylinder, and it's narrow enough to still have positive g loads at the inverted position, then it must necessarily pull more than 2 g at the bottom. That bit would not be safe with almost all types of airplane. I don't think pilots do it this way in practice, they will rather follow a curve that has a much tighter radius on top than at the bottom. (Which means it's arguably not a barrel roll anymore, but will still be perceived as such.)

again check out the mindboggling video Wayne posted above!

@Fattie Bob Hoover is widely regarded as perhaps the best stick-and-rudder man there has ever been.

@WayneConrad - you rock

@DavidK the WW2 training film is really fantastic. So carefully made and executed.

I guess it means the pilot managed to provide a net 1 g of
acceleration upwards, so the body and any other object experienced the
usual head-to-toe force.
But to achieve that (1G) acceleration it should flew with 2 g of
acceleration (to compensate the gravity first). She said the flight
path was pretty straight.

No, you are mistaken here, unless I am misunderstanding your meaning, which is possible.  (These concepts are kind of "squirrely" -- the net acceleration on the aircraft is indeed 2 G if the aircraft is inverted and the occupants are experiencing the same sensations that they'd feel in normal upright flight, but there's no "compensation" going on.  Read on for more.)
The net acceleration in "usual" life is actually 0 G.
For example, standing on the ground, the ground is pushing up on our feet with a 1 G upward force, and gravity is pulling down on our body with a 1 G downward force.
Flying straight-and-level, the aircraft wings are generating a 1 G upward force, which is transferred through the aircraft structure, and ultimately through the seats, to our body, while gravity is pulling down on our body with a 1 G downward force.
In what we call "0 G" flight or "weightlessness", the aircraft wings are unloaded (i.e. the angle-of-attack is reduced) so that they generate 0 G of force.  Gravity still pulls down on us with 1 G of downward force, so the net force is 1 G downward.
To experience the sensations of normal upright flight while inverted, the aircraft must generate 1 G of force in the "upwards" direction in the aircraft's own reference frame.  This means the aircraft must generate 1 G of earthward force, i.e. toward the ground.  Meanwhile, gravity is also pulling down with 1 G of earthward force.  The net force is 2 G, acting in the earthward direction.  So the lift force generated by the aircraft is not compensating for gravity, rather it is adding to the effect of gravity, in terms of the net acceleration experienced by the aircraft.
You can see why this would not be sustainable for long-- the flight path is going to be curving earthward very rapidly.  Even if the segment starts with the aircraft's nose pointed well skyward, it will soon be pointed steeply downward.
However, it's been pointed out in a comment that even a slight mild positive G-loading generated by the aircraft will produce sensations that might be hard to distinguish from the normal full positive G-loading of 1 G, especially for an inexperienced passenger.  Still, even if the aircraft was producing a positive G-loading that was just barely above zero, the net acceleration on the aircraft will still be slightly more than 1 G earthwards, so even in this case the maneuver is only sustainable for a relative brief interval.  It's pretty clear that your informant is simply not reliable-- either she is not remembering her perceptions clearly, or she was too overwhelmed by the newness and strangeness of the experience to clearly perceive what was really going on.  (Especially this part: "She said the flight path was pretty straight.")
"Reading between the lines" of this answer, I hope you can see that the only force we actually feel in flight is due to the real aerodynamic and thrust force generated by the aircraft.  We don't "feel" the force or acceleration component that is due to the pull of gravity, because gravity acts "from within" and exerts an equal pull (per unit mass) on every particle of our body, creating no stresses or strains, and thus no sensations.
Disclaimer: this answer is written from a Newtonian perspective, treating gravity as an actual force.
Disclaimer: this answer has treated G-loading as a measure of force as well as acceleration.  Strictly speaking, G-loading is not actually a measure of force, but a measure of force per unit mass, which is the same thing as acceleration.

Very interesting answer. Thanks for taking the time to write it.

the net acceleration on the aircraft is indeed 2 G in "weightless" flight - Should be 1 G downward for "weightless", airplane accelerating downward around the freefalling occupants.  Or 2 G downward for "feels normal" inverted flight (occupants feel 1G "downward" in the reference frame of the inverted aircraft).  This is what the rest of your answer explained, so I assume that early sentence is a typo or editing error.

@PeterCordes -- yes, I slipped a cog somewhere, edited, thanks

Peter Kämpf has the right idea here.  It sounds very much like the pilot is doing positive G or so-called “Gentlemen’s aerobatics”.  Aileron rolls, barrel rolls, and even some variants of inside loops are all positive G maneuvers.
In regards to your comments, yes, an aircraft inverted, and experiencing a positive G loading would have to be experiencing an acceleration towards the Earth greater than 9.81 m/sec^2 in order for the air frame and pilot to experience this. This kind of flight regime and loading is possible for a limited period of time while inverted. In sustained, level, inverted flight or in an outside maneuver, the pilot would experience a negative G loading.  Slow rolls and outside loops are some examples of negative G aerobatic maneuvers.

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text:
How can a designer maximize the glide ratio of an aircraft with low aspect ratio wings, other than by increasing the aspect ratio?
As I understand it, the gliding range is given by a lift to drag ratio—hence to increase the gliding range, I would want to increase the lift and reduce the drag as much as possible.
To that end, apparently, it seems that high AR fits both the bills. However, does that mean that I will not able to enhance the gliding range with low AR wings at all?

Sure you can.  Shave off some of the wing chord to make them into higher aspect wings.  The increase in wing loading won't decrease your gliding range in still air, and it will actually increase the gliding range if you are gliding into a strong headwind.  Also, make like hawk and get some slotted wingtips, ideally ones you can close at will when you need to fly fast.

Highly related (if not a dupe): Relation between drag and aspect ratio

Does this answer your question? What dictates the aspect ratio of an aircraft's wing?

With a given wing span, everything that reduces drag will increase the gliding range. Aspect ratio is less of a factor here because induced drag depends on wing span, not aspect ratio. But increasing the aspect ratio with a given wing span will reduce the wetted surface and, consequently, friction drag.
What can be done to increase gliding range outside of increasing span?

retract the landing gear,
avoid struts or bracing wires,
minimize friction drag by allowing for more laminar flow and minimum surface area.
minimize pressure drag by streamlining everything

Generally, a low aspect ratio shifts the polar point of best gliding range to lower lift coefficients, so reducing all that parasitic drag becomes more important the stubbier the wings are.
The most effective way, however, to increase gliding range is indeed a larger wing span if all other parameters are held constant. Increasing span while keeping wing area constant will:

increase aspect ratio and reduce wing chord,
reduce induced drag at the same lift coefficient,
and, as a consequence, lower the optimum gliding speed, so the parasitic drag is lower as well.

I think you should describe the effect of low and high aspect for the same wing area.

@JohnK it has already been done in this answer

No that answer also talks about different ARs with the same span, so one has less area than the other.  I'm talking about two wings of the same area, say because I want to consider two cases with the same wing loading in order to have the same stall speed.  A 1-26 has 160 sqf and 1-23 has 165 sqf, more or less the same, AR of 10 and 17 respectively.  L/D of 23 and 31 respectively.  For two wings of the same area, the higher AR does better even tho wetted area is the same. I believe the answer is related to lower tip losses and the higher AR is moving a larger package of air a smaller amount.

@JohnK: The higher AR does better because wingspan is larger. 16 m versus 12.2 m - that is about the ratio if their respective L/Ds, too.


text:
Can yaw develop a spiral dive?
I'm wondering if the 'prolonged' effect of using yaw causes a spiral dive. I understand that a spiral dive is caused by an excessive angle of bank, but can it also be created by an excessive amount of yaw? A spiral dive has the characteristics of a rapid speed increase, rapid loss of altitude and unstalled wings. An aircraft that is yawing too one side for a long time does all of these things, so shouldn't it the prolonged effect of yaw be a spiral dive?
Also, when the aircraft does a skidding turn, why doesn't the aircraft develop a spiral dive? Why is it only when an aircraft does a slipping turn, that the aircraft may only enter a spiral dive?
Thanks.

Where did you get the idea that there was some association between sideslipping and entering a spiral dive?  If you've encountered this idea in flight training materials, it would be helpful if you'd add a citation to your question.

Your first paragraph asks if it is true, and your second paragraph assumes it to be true.  Anyway, aircraft stability will determine any tendency towards uncorrected yaw developing into a spiral.  I don't really think there is a universally correct answer to the question.

I'm wondering if the 'prolonged' effect of using yaw causes a spiral
dive. I understand that a spiral dive is caused by an excessive angle
of bank, but can it also be created by an excessive amount of yaw?

If by "yaw" you mean the aircraft's yaw rate, then we can observe the following--

In a low-airspeed constant-altitude turn where the bank angle is rather steep-- say 45 degrees-- the aircraft's yaw rotation rate is high, but the aircraft is not in a spiral dive.  (A sailplane circling in a thermal updraft is another example of an aircraft with a high rotation rate, that is not in a spiral dive.)  But in this situation, the pilot may be holding some rolling-out input with the ailerons to prevent the bank angle from increasing-- see below for more on this.

Even in fully "coordinated" flight (aircraft is aligned with the instantaneous direction of the flight path and therefore is pointing directly into the "relative wind"), a high yaw rotation rate creates a difference in airspeed between the two wingtips which tends to make the bank angle increase.  Unless the pilot makes a roll input with the ailerons to counteract this, this can tend to make the aircraft enter a spiral dive.

If we are flying along in wings-level flight and then we stomp on one of the rudder pedals to establish a high yaw rotation rate, we'll end up in a skidding turn which can indeed lead to a spiral dive.  See below for more.

Also, when the aircraft does a skidding turn, why doesn't the aircraft
develop a spiral dive? Why is it only when an aircraft does a slipping
turn, that the aircraft may only enter a spiral dive?

This is a misconception.  In a typical general aviation plane, if you trim for level flight, take your hands off the control yoke, and then hold full rudder, you'll end up rolling into a rather steep bank in the direction of your rudder input.  The roll is driven not only by the difference in airspeed between the two wingtips that we noted above, but also by the sideways airflow interacting with the dihedral geometry of the wing (including related effects such as the dihedral-like effect generated by a high-wing configuration.)  You'll find yourself in a spiral dive in the direction of your rudder input, with the slip-skid ball displaced toward the outside or high side of the turn.  In other words, you'll be in a skidding spiral dive.
It's true that, in the absence of pilot rudder input, a spiral dive will typically be associated with a small amount of a sideslip-- but this is true of level turns as well.   Even if sideslip is present in a spiral dive, the sideslip is not really what is driving the spiral dive.  Adding inside rudder to eliminate the sideslip will typically make the bank angle get steeper, which will make the dive angle get steeper.
This answer is built on the concept that a steep bank angle is a key element of a spiral dive.  If you are trying to ask if a spiral dive can develop due to an excessive yaw rate while the bank angle stays shallow, the answer would generally be "no".  If you enter a severe skidding turn while using the ailerons as needed to keep the bank angle shallow, the extra drag from flying sideways through the air will tend to increase the sink rate, but not to the extent that the maneuver would generally be described as a "spiral dive".  But don't forget that a skidding turn can be an invitation to a spin entry!  That's the classic "trap" in the landing pattern-- the pilot perceives a need to increase the turn rate but is hesitant to  increase the bank angle, so he (perhaps unconsciously) tries to increase the turn rate by using extra rudder instead, but this tends to lower the nose, tempting the pilot to move the stick or yoke excessively aft-- and the plane stalls and spins.

Ok, so in short: yaw can develop a spiral dive?

@user14397644 yaw rate is what drives the increase in bank angle, so yes.

Regarding the yaw/roll coupling in typical GA aircraft, it's actually possible to control the aircraft for long periods without the yoke: I've flown entire cruise legs using gentle rudder for directional control and trim/power for pitch control. It's a good exercise to develop a new dimension of feel for the airplane.

@TypeIA -- it's even  more fun to practice a series of reversing turns with bank angles up to 30 degrees or more, without touching the yoke.  It really gives one a feeling for how the nose tends to rise whenever the bank angle rapidly decreases, due to the aircraft retaining excess airspeed (and therefore excess lift) from the turn, until the excess airspeed has time to bleed off.

(ctd)  Sometimes the pilot is forced to use roll (accomplished in this case via slip-roll coupling, with the rudder pedals) to control pitch-- e.g., increase the  bank angle to bring the nose down, or decrease bank angle to raise the nose.

It must be remembered that the wing, at any given moment, is generating more force than any other part of the airplane, including the engine, by far.
Slipping turns are much more prone to create a spiral dive because the critical ingredients are already there:  loss of lift from the steeper bank and a stronger force pulling the plane sideways.  Because of the bank, the horizontal empennage, including the elevator, will begin to pull the plane into a descending spiral.  Pilots, seeking to stop the descent (especially when disoriented by lack of VFR conditions), will tighten the spiral by pulling on the elevator more.
Vertical lift is much less affected by the smaller bank of the skidding turn, so it is less known to create a spiral (but more known to create a deadly tip stall).
The key element of a "skid" input (rudder) is the ability of the rudder to roll the airplane.  High wing trainers have a lot of area underneath the CG and must be "helped" with the ailerons to maintain bank angle in a turn.  Without aileron input they will have a greater tendency to "skid" (inclinometer ball to the outside).
Low wing aircraft, with more area above the CG, and dihedral, can be "snap rolled" with hard rudder alone.
Gliders will also have a greater tendency to roll with rudder owing to the wing tip speed differential of their long wings.
So, one can see it is possible to design an airplane to do a coordinated roll/yaw turn with rudder alone.  (If it rolls enough, it is no longer skidding!).
A skidding turn does create a lot of drag, which can start a downward spiral path due to loss of airspeed, but it is the wing and horizontal surfaces in a bank that are the real culprits to this hair-raising scenario.
Fortunately, one can recover by relaxing elevator, cutting power, and rolling to level before pulling out.

Re "Slipping turns are much more prone to create a spiral dive because the critical ingredients are already there: loss of lift from the steeper bank and a stronger force pulling the plane sideways."-- I question this.  Either a slipping turn or a skidding turn can be flown at any bank angle.  In a slipping turn, dihedral will generate a roll torque towards wings-level, while in a skidding turn, dihedral will generate a roll torque away from wings-level.

(Also, for a given bank angle, the wing's lift vector is larger when the plane is skidding than when it is coordinated, and larger when it is coordinated than when it is slipping.  For a given turn rate at a given airspeed, on the other hand, the bank angle and the wing's lift vector are both least in the skidding turn and greatest in the slipping turn.)

Your statements about the effects of high-wing or low-wing configurations don't seem accurate to me either.  Oh well, the topics of slips and skids have proven fertile ground for many competing answers in the past as well--

But in general, skidding turns are considered an invitation to a stall-spin accident.  Accidentally allowing some slip in a turn is generally not considered dangerous, though your answer is suggesting that it is an invitation to a spiral dive--??

@quiet flyer A turn at a given rate will have a steeper bank on the slip.  Vertical CG (in relation to side area) does play a role in roll rate from rudder application (a dihedral "effect").  While coordinated is best, a spiral with wings unstalled would be "safer", than a skid induced tip stall, which can be sudden, especially low. The tip stall recovery technique for the DC3 involved reverse ailerons, which would not be good if the tip wasn't stalled.  These thoughts made me very interested in slats.

I guess it all depends on what variables we hold constant.  In the real world, for manned aircraft (as opposed to model airplanes), seems to me a spiral dive is normally only a hazard encountered when instrument flying, and I'm having a hard time imaging a scenario where a tendency to allow sideslip while turning would make a spiral dive more likely to develop.  Because of the rolling-out torque that is normally created by a sideslip.  But yes you are right, for a given turn rate, more bank is required if the plane is slipping than if it is not.  On the other hand, for a given bank angle..

I'm also having trouble reconciling your suggestion that high-wing aircraft tend to roll out of turns, w/ your suggestion that low-wing aircraft tend to exhibit a stronger roll response to rudder inputs.  In truth, enhanced roll stabilty and enhanced coupling between slip and roll are both consequences of "effective dihedral".  "Effective dihedral" can be created by a high-wing geometry, or by sweep, as well as by actual dihedral.  Anyway I can't agree w/ the suggestion that it is easier to use the rudder to control roll in low-wing planes than in high-wing planes.

@quietflyer you are correct that trying to generalize "high or low wing" is not the best way to describe it.  More accurate is side area in relation to CG, which can be used to create an "anhedral or dihedral effect"  which either prevents or couples a roll into a yaw input.  Ironically, the more it rolls, to less it is skidding.  Years ago an instructor showed me how to turn a R/C plane with rudder.  Before that I just banked and yanked.  It was a high wing trainer with a lot of dihedral, so I see what you are saying.


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How can L/D Max be both Best Glide speed and Max Range speed?
According to the book "Aerodynamics for Naval Aviators", best glide speed AND max range speed for propeller aircraft both occur at L/D max.
This is a little confusing because in an A36 Bonanza Vbg is 110 KIAS and Vmax range has been experimentally calculated by several sources to be about 125 KIAS.
So my explanation for how both of these are L/D Max is that L/D Max is a specific AOA, not a speed.
However, this brings me to my question.  Obviously Vbg is power off and Vmax range is power on. Assuming that L/D max is around 6 degrees AOA, with no power the airplane is descending at about 700-1200fpm (depending on prop pitch and other configuration items) at a speed of 110 KIAS.  At max range speed the airplane is level at 125 KIAS and presumably also at an AOA of approx 6 degrees.
Something seems off here.  Here is what I’m having a hard time correlating:  I teach that trim controls AOA, not speed.   If the airplane is trimmed for an AOA of 6 degrees, adding or removing power should only change the rate of climb/descent, not the airspeed.  Adding power gradually from the power off glide config at 110KIAS will eventually result in
level flight AT 110 KIAS.  If we wanted 125 KIAS we would have to add power while trimming down (lowering AOA) which means we are not at the same L/D Max AOA at Vmaxxrange that we were at Vbg.
I feel like I’m missing something here, so either one of my assumptions is incorrect, or there is another puzzle piece I’m missing that explains this. If we have to retrim for 125 KIAS level with added power, how can we be at the same l/D max AOA?

Ok.  I think I figured it out.  Please correct if there are inaccuracies.  The statement that max range AND max glide both occur at L/D max does NOT logically require that those 2 speeds or AOAs are the same.  L/d Max for gliding flight and l/d max for powered level flight are different angles of attack.  The difference is the drag curves.   A gliding aircraft has less lift than a level one. Similarly, a windmilling prop has different drag characteristics than a thrust-producing prop.  So since the curves differ, l/d max will occur at a higher AOA for gliding flight than powered flight.

I'm going to answer with the caveat that I don't have as deep of an understanding of this as I would like, and someone smarter than me may be able to correct or expound on my response.
Best Glide is a purely aerodynamic concept. There is no powerplant involved, so you're simply converting potential energy (altitude) into forward motion, at a rate and L/D ratio that is driven by the angle of attack. It's a simpler concept.
When you add a powerplant into the equation, it gets a lot more complicated. The powerplant itself has modes where it's more efficiently converting fuel energy into thrust, and modes where it's less efficiently doing so. Max Range, therefore, depends not only on aerodynamics, but also how efficient the powerplant is. When you add powerplant dynamics into the equation, you might settle on a best range speed that is different than the purely aerodynamic best glide range speed.
Simply put...

Best Glide is where L/Dmax occurs.
Max Range occurs when (FuelBurn / GroundSpeed) is the lowest.

On a related, but slightly different, note: Piston airplanes and Turbo(fans/jets) behave differently when it comes to producing power at different airspeeds. It's why we generally measure turbofan engines in thrust, and piston engines in horsepower. This difference is why Max Range and Maximum Endurance speeds correlate with Best Climb Angle and Best Glide speeds differently, when you compare pistons to jets.
I hope this helps.

What an understatement - this answer does show deep understanding. +1.

Username checks out :D

In addition to fuel-to-thrust efficiency, there's also the effect of thrust line in relation to centers of gravity and lift. Changing power setting (usually) changes trim, which in turn changes the efficiency of the whole setup.

@bianfable Yes, I approve.

@RobertDiGiovanni I fail to see how your comment addresses mine.

@Jpe61 I removed it.  The "efficiency of the whole setup" is indeed fascinating.

Best L/D is a purely airframe concept, whereas best range is a wholistic system concept.
(This answer builds on @LDMax's answer)
Analysis
Airframe's best L/D
This is the exact point on the glider polar where any faster or slower results in a shorter glide range (in still air).
However, sometimes that peak is very shallow, that is to say that accelerating or decelerating a bit doesn't dramatically change the glide ratio. In fact, that's a desirable characteristic because it means the pilot isn't trying to balance on the tip of a needle. The takeaway is that speeds above and below best L/D might also return similar, albeit not mathematically maximal, glide ratios.
Interpreting the below graph in this context is easy. The glide slope is geometrically the slope of a line starting from the origin and intersecting the polar curve. The shallower the line, the better the L/D. In the below example you can see that the curve is relatively straight after the peak and so FOR THAT AIRPLANE flying 10% faster doesn't dramatically change the L/D.

Engine's peak efficiency
An Otto cycle engine makes peak efficiency when the throttle body is wide open and there is sufficient load to keep the engine from revving too fast. This is easy to understand, since the major efficiency loss in an engine is pumping loss. If the throttle is closed, then airflow is restricted and the engine has to work harder just to breath (imagine trying to breath through a straw). If the engine is revving high, then it's pumping a lot of air (imagine breathing heavily for minutes on end).

Note that the difference in peak efficiency can be substantial. Going just a little bit faster can increase the engine's efficiency by 10%.
Propeller peak efficiency
A propeller's efficiency is a balance between slipping too much and having too much drag due to high speeds. It starts low, peaks high, and comes back very low. Constant speed props help match the best propeller RPM to the best engine RPM, but there's only so much they can do.

Once again, note that the difference in peak propeller efficiency can be substantial. Going just a little bit faster can increase the propeller's efficiency by 10%.
Conclusion
The best L/D is a question which involved only one subject, which is the airframe in a gliding configuration.
The max range is the amalgam of several systems, which have somewhat contrary requirements. So you can be overall by going a little faster, which doesn't substantially increase airframe drag but does substantially decrease engine and propeller losses.
In the case of a Bonanza, it works out to a steady-state engine and airspeed setting. However, in the case of a self-launching glider with retractable engine pylon, the best range is attained through a "sawtooth" flight pattern, which is a combination of pure glide phases and maximal climb stage. This is because 1) with the engine stowed the drag of the propeller and engine is removed and 2) when the engine is deployed it is only run at a maximally efficient climb setting.

Re -- "in the case of a self-launching glider with retractable engine pylon, the best range is attained through a "sawtooth" flight pattern, which is a combination of pure glide phases and maximal climb stage. " --I'm pretty sure that there was an ASE question about this sometime in the last few months.  Don't recall whether it was asking specifically about motorgliders.  It might be worth your time to search for this question and post an answer, or maybe you already did answer it.

This is the question https://aviation.stackexchange.com/questions/79854/can-gliding-your-plane-save-fuel/79956#79956

The short answer is that parasite drag changes because of the slipstream effect of an operating engine (in the case of a propeller at the front), thus changing the L/D max speed.

The general "textbook" formula to show calculate the max range speed (at least the one I was taught), goes more or less as follows:
For a piston engine, fuel consumption is roughly proportional to engine power:
[1] fuel_consumption_per_hour = constant * power

Hence
fuel_consumption_per_mile = constant * power / ground_speed

Assuming no wind, gs = tas, power is thrust * tas. In steady level flight, we get power = drag * tas.
Plugging it back:
fuel_consumption_per_mile = constant * drag

Which is minimum for lowest drag. For a given weight, assuming lift = weight, we can get:
D = (D/L) * W

So in order to minimize the fuel consumption per mile, we need to maximize the L/D ratio.
All of this hinges around the assumption [1], which works for an ideal piston engine with perfect efficiency.
Furthermore, as you pointed out, the relationship between angle of attack (and airspeed), and L/D ratio normally is affected by the engine RPM because of aerodynamic interference. (Also in certain aircrafts, thrust vectoring effect will significantly affect the tail lift, making the entire relationship even more complicated).
Conceptually, one would have to:

Select optimal L/D ratio
Compute AOA and Airspeed (weight dependent) that results in the selected L/D ratio
Compute engine power required for steady flight
Re-enter at 2 with new engine power setting, and repeat.

In practice an approximate answer is always fine. In your case an error of 15 kts doesn't seem to terrible.

The fact that less lift is required in gliding flight than in level flight is a bit of red herring. The difference is tiny and can be accommodated by flying at the same angle-of-attack and L/D ratio and Cl/Cd ratio but at a (indistinguishably?) slower airspeed. For all practical purposes, for reasonably efficient airframes, this effect is so tiny that any given angle-of-attack corresponds to the same airspeed in gliding flight as in horizontal powered flight.  But you are right about prop drag changing the drag curve in an unpowered glide, which would shift the best L/D ratio to a lower airspeed. Also, prop and engine efficiency considerations could cause the actual best range speed to be different than the speed for best L/D ratio.
At the end of the day, due to prop and engine efficiency considerations, the idea that the still-air best range occurs at the speed for the best L/D ratio is only an approximation.  On the other hand, it's always true that the still-air best glide angle occurs at the airspeed and angle-of-attack that yield the best L/D ratio for the existing aircraft configuration (which may include windmilling prop, stopped prop, feathered prop, idling prop, etc...)


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Why does increasing speed in a turn not change load factor?
From what I've read, load factor (n) is the forces of lift (F) divided by weight (W), so the equation would be n=F/W. In a question about a constant bank angle turn, an increase in airspeed will decrease the rate of turn, which I understand, but will not affect the load factor. If airspeed increases, doesn't lift increase, which should increase* the load factor? I'm missing something here about the calculation of the forces of lift, but I'm not sure what.
*edit

When you increase airspeed but change nothing else, lift will increase. But if you apply more downward pitch to stay at the same altitude, your lift stays the same. Maybe whatever question you refer to assumes no change in altitude.

If the weight of the aircraft is unchanged, why wouldn't the lift be? Increasing the speed increases the dynamic pressure, but you must reduce the $C_l$ proportionally so that the lift remains unchanged.

Is the flight path constrained to remain horizontal?  (See my answer for context)

You need to think about what are the constraints and what are the variables.  If bank angle is a "given" (i.e. constrained) but turn radius or rate are not, and the airplane is constrained to remain in horizontal flight (not climbing, and especially not "pulling G's" to curve the flight path ever-more-steeply-upward in an approximation of the start of a loop), well, then for any given bank angle, why should increasing the airspeed increase the load factor?
Keep in mind that for any given bank angle, every extra knot of airspeed mandates a decrease in the angle-of-attack of the wing in order that the flight path remain horizontal rather than starting to curve upwards.  So why should the load factor increase?
At the end of the day, if the only "givens" are bank angle and the fact that the flight path must remain horizontal, then the flight dynamics don't "care" about airspeed at all as far as load factor is concerned.  The load factor will always equal (1/ (cosine (bank angle))).
On the other hand, if turn radius or turn rate is included as one of the constraints, and bank angle is not, then everything changes: an increase in speed will mandate an increase in bank angle and load factor.

If airspeed increases, doesn't lift increase, which should increase
the load factor? I'm missing something here about the calculation of
the forces of lift, but I'm not sure what.

You are missing that the angle-of-attack is not constrained to be constant, but the flight path is constrained to be horizontal, and that's why increasing the airspeed doesn't increase the lift force.
P.S. your question didn't actually state that the flight path was constrained to remain horizontal.  But in the absence of information to the contrary, that's pretty much presumed (though it really should be stated explicitly).  If the airplane is allowed to start entering a loop as soon as we increase the power, then situation becomes immensely more complex, and since we aren't given any explicit constraints on angle-of-attack (e.g. a-o-a is presumed to remain constant?), we're basically left with an unsolvable (insufficiently constrained) problem.

If I understand correctly - because the flight path is constrained increasing the airspeed actually decreases the angle attack keeping lift relatively the same? And because the aircraft is moving faster, less of change in the angle of attack (compared to the slower speed) is required to generate the required lift to maintain level flight in the turn? (which is why it is easier to overload the plane at higher airspeeds)

What is preventing changing the value of the other variables in the lift equation in order to generate the greater Lift required?

@RyanBW -- re your last comment, I'm not sure how to explain it any better than my answer did.  It's a "given" in the original question that the airspeed has increased, while bank angle has remained constant.  It's arguably implied that the flight path is horizontal, not partially resembling a loop.  So why does the change in airspeed require more lift?

@RyanBW-- re your first comment, it appears that you are kind of asking a new question.  It appears that you are asking whether the delta (change in) AoA that is required as we go from wings-level flight to a constant-altitude turn at some specfied bank angle, at a constant airspeed, ends up being higher when the airspeed is low than when the airspeed is high.  If that is what you are now asking in the comment, it seems to be something not fully addressed in the present question or answers.  Maybe you should ask it!

@RyanBW -- because the final load factor ends up being the same in both cases.  And at any given airspeed, for any given aircraft, if AoA is constrained to a specific value, we can calculate the load factor, and it will be independent of bank angle (but we cannot now constrain the flight path to be horizontal, we may be entering a tilted loop!)  And yes if you change the AoA by X degrees there will be more change in load factor if you are flying at a high constant airspeed than at a low constant airspeed.  But answering the original question doesn't seem to be dependent on that fact.

It's just like a plane in straight and level flight.  If you increase speed by increasing thrust, Angle of Attack must be reduced to maintain level flight (not climb).
Same thing for 30 degree bank angle turn.

Radius of turn is V$^2$ knots/11.26×tan bank angle in feet

Increasing speed increases radius.

Load factor does not increase

For level flight vertical lift component = Weight.
Load factor at 30 degrees bank is 1/cosine 30 degrees = 1.15 G.
The $cosine$ of the lift component is the vertical lift component.  The $sine$ of the lift component is the centripetal force into the turn.
Since these do not change, load factor is the same, and the rate of turn decreases.

Rate of turn is 1091×tangent bank angle/V knots in degrees/second


text:
Does CG change airspeed in a descent or glide?
If the CG of an airplane moves forward, would it fly a faster or slower airspeed in a descent?
Usually during a descent or landing, you'd use pitch to change airspeed and power to control descent rate. With full nose-up trim, power off, a plane I fly frequently used to fly at a certain speed before some avionics work was done, but now sits at 5 knots faster with the same configuration, same weight (me). This is in a descent to landing. I assumed initially that maybe the CG moved forward a bit and the speed increased due to the plane wanting to be more nose-down with the same configuration and trim. However, I remembered that a forward CG usually decreases performance (and thus speed?) due to a larger tail down force needed to counteract the forward CG, more wing loading and more drag. These two ideas seem to contradict one another! I also don't know if a power off descent vs level flight changes the fact that a more forward CG, increases drag and decreases airspeed.
Thank you for anyone's input.

Moving the CG forward will result in a negative $∆C_m$ (nose-down moment) for the wing, with minor effect on the tailplane $C_m$, which depends on the direction of its loading. So yes, if the original trim setting is preserved, a net negative moment will be produced at the original airspeed, meaning that the aircraft will now be trimmed for a higher airspeed.
It is true that forward CG comes with a small drag penalty (check out this question for more information on that). However, that doesn't affect the trim speed at all.
When a stable aircraft is trimmed for a given AoA (and therefore IAS), it will maintain that by producing an appropriate pitching moment to counter any deviations. When you increase the thrust, the aircraft pitches up to maintain the IAS. Likewise, a reduction in power or increase in drag will cause the aircraft to pitch down to maintain the IAS. Drag has no effect on the trim speed; increased drag will only cause the aircraft to have a greater descent angle with the same original speed.

Thank you, that makes a lot of sense. So basically, in level flight, with a more forward CG, drag is increased and performance is reduced because there is more fuel burn/thrust needed to maintain the same speed without altitude loss. The drag energy cost is replenished by fuel burn. Whereas in a descent with power off, even if forward CG increases drag slightly, the energy cost there is being taken from altitude/potential energy. What's not changed is the relationship between trimmed AoA and IAS and the stable aircraft's "desire" to maintain that AoA

If the CG of an airplane moves forward, would it fly a faster or
slower airspeed in a descent?

Hang gliders and "trikes" (powered hang gliders with seats and wheels) use CG-shift as the sole means of pitch (and roll) control.  Anyone who has flown such aircraft will tell you that the answer to your question is "faster".

However, I remembered that a forward CG usually decreases performance
(and thus speed?) due to a larger tail down force needed to counteract
the forward CG, more wing loading and more drag

If the airplane is re-trimmed to continue to fly at some given airspeed at the original power setting after the forward CG shift, drag will generally be increased due to the increased tail downforce.  So the climb rate will be decreased, or the sink rate will be increased, depending on the power setting.  Similarly, if the airplane is re-trimmed to continue to fly horizontally at some given power setting after the forward CG shift (or to maintain some given climb rate or sink rate), the airspeed will generally be decreased due to the increased drag associated with the increased tail downforce.
But if you leave the elevator trim fixed as you move the CG forward, the tail isn't going to magically create more down force so as to decrease the airspeed.  Rather, you'll end up flying at a lower angle-of-attack, and so you'll see an increase in airspeed.1  If you are flying on "front side of the drag curve", as is most often the case, this will cause an increase in sink rate (or a decrease in climb rate).  If you are flying on the "back side of the drag curve", as is the case in "slow flight", this will cause a decrease in sink rate (or an increase in climb rate).

I also don't know if a power off descent vs level flight changes the
fact that a more forward CG, increases drag and decreases airspeed.

There's no fundamental difference in this regard between a power-off descent and level flight.  It's just a matter of what are you holding constant as you change the C.G. position-- elevator trim position, or altitude, or airspeed?  See above for more.
Footnotes:

A careful analysis of this situation shows that the tail downforce has in fact been increased.  But the result is not a decrease in airspeed.

From what I know if the CG moves forward it would result in a 'nose heavy' situation. This would cause the aircraft to go into a dive. In other words it would have a negative AOA which you wouldn't want during landing. If the CG moves aft it would cause a 'tail heavy' situation and as a result it would cause a positive AOA such as your nose will pitch up.
In order to compensate these changes the elevator is at your disposal or sometimes pilot would normally lean/push back or forward (depending on the condition) their seat especially if your in a light body aircraft. The decent rate can be manipulated by increasing the lift with the aid of flaps.
It must be noted that any movement of the CG can also affect the center of lift on the wings. For example, if the AOA is increased (pitch up) the COP moves forward which will in fact cause a reduction in lift. Therefore you'd need to add more power to compensate. All in all, what you'd want to do is perform weight and balance procedure to correct the aircrafts CG if you think it was disturbed.
Thanks

Welcome to Aviation Stack Exchange. The question was the affect of the center of gravity changing on airspeed, not on Angle Of Attack. Please try to make sure you are answering the question being ask.

Welcome to Aviation Stack Exchange. The question was the affect of the center of gravity changing on airspeed, not on Angle Of Attack. Please try to make sure you are answering the question being ask

Well, if the CG is moved forward and during a decent it would increase your decent rate because the COP will also tend to move forward as a result you will lose lift if correction is not made. Gliding? I don't know(I haven't seen anyone in an aircraft want to glide) but in crusing or level flight you would want to compensate this nose-heavy situation by increasing your thrust/power

Moderate forward movements of the CG do not place the wing at a negative AoA.

@Jim  -- changes in AoA are the fundamental cause of changes in airspeed. See https://aviation.stackexchange.com/a/100452/34686.

@LudwigT - re "(I haven't seen anyone in an aircraft want to glide)"- a) sailplanes (gliders) are aircraft. So are hang gliders and paragliders. b) assuming you mean "powered airplane"- you've never seen a landing approach at idle power? You've never seen an airplane shut down the engine for a mountain wave soaring adventure? Both are commonplace. c) In an airplane, your main compensation for a forward CG condition will be aft stick/yoke (up elevator) (can accomplish same with trim). Yes, some extra power will be needed to generate the same airspeed in level flight, but this is secondary.

@LudwigT -- "For example, if the AOA is increased (pitch up)... which will in fact cause a reduction in lift."  -- ??? Needs some improvement.


text:
Flaps type for Beech Bonanza V-35B
I'm writing a thesis on airplanes' flaps and I was analizing the flaps on the Beech Bonanza V-35B: on the book "The wind and beyond - Volume II" published by NASA on page 375 states that this plane uses "single slotted flaps".

However looking closely at some pictures I think it uses single slotted fowler flap, since they seem to slide back towards the aft part of the wing.

In general I think that some books use the term "single slotted flap" to describe the "single slotted FOWLER flap", because the latter is far more used, but this creates some confusion.
What kind of flap is this airplane exactly using? and can you provide me a decent reference (not wikipedia or something similar) in support?
Many thanks

Feel free to roll back edit if you wish-- I was assuming your were (incorrectly) punctuating to denote plural, but I guess another view is that the attribute of "type" is something that is possessed by the flap--

Yes, the single slotted flap on the Bonanza does have some Fowler type rearward motion. There is no strict convention on what defines a Fowler Flap.
Here is an article which discusses the question of Fowler flaps in light General Aviation aircraft. FLAP BIAS ... WHY? WHEN?

“As we were writing this page, we consulted a few friends. The well-known aviation writer Peter Bedell (a Baron owner) noted: "[Whether a light aircraft's flap is truly a Fowler or not] is a gray area...For example: The Baron's flaps do go aft and they do increase wing area as Fowler flaps do. This is done by a clever track/roller system in the wing and flaps themselves that guides them aft and then down as Fowlers should. I believe the Bonanza through King Air use the same system. I don't know how much aft travel there must be to define a "Fowler flap". I was led to believe that it was any flap that increases wing area.”
"Technically, the King Air's, Baron’s, Travel Air's, Bonanza’s, and Cessna 182/172’s flaps can be considered Fowlers…but the typical effect of the Fowler that design engineers of heavy airplanes try to achieve (that being: significant chord extension with added camber) is not being achieved in these light general aviation designs…and need not be.”

Thank you for the answer! On many of the books I checked the fowler flap is described as a device that both increase the camber of the airfoil and the wing area. For example, Anderson in "Aircraft performance and design" writes: "The Fowler flap, when deployed not only deflects downwards, hence increasing the effective camber, but also translates to the trailing edge of the airfoil, hence increasing the exposed wing area with a further increase in lift." So I think that increasing the wing area is a feature of the fowler flap (in all it's variants), otherwise it'd be a "simple" slotted flap.


text:
Does the reduced aerodynamic damping at higher altitudes affect the Vno (maneuvering speed) of an airplane?
Since reduced aerodynamic damping at higher altitudes reduces the needed control input for maneuvers, is this just about the required force to move the controls or also about the speed of the resulting motion of the aircraft? If the latter, would full deflection of the controls at high altitude result in higher g-forces on the airplane for the same indicated airspeed, and because of this a lower Vno? Or is the effect of full deflection of the controls on the airplane actually the same at high altitude as it is at low altitude, regardless of aerodynamic damping?

Interesting.  TAS will be higher for a given IAS at altitude. Turn radii will be greater.  G force loads and control deflections will be similar (up to critical Mach values).  Less damping (thinner air) may lead one to err on the side of caution changing control settings due to potential overshoot while changing, but the Vno indicated may be similar, but limited to a Mach (TAS) value.  (Airliner pilots please chime in).


text:
How would IAS change with altitude for an unpowered parafoil?
Suppose an unpowered parafoil (e.g. a ram air parachute or paraglider) is descending (gliding) in a standard atmosphere. Suppose a pitot tube is positioned to always point into the relative wind (e.g. hanging by a string with fletching) and is recording IAS.

I know that the TAS (true air speed) is expected to be higher at higher altitudes due to the reduced drag. How is the IAS (indicated air speed) expected to change with altitude?

Would the answer be different for a fixed wing glider that has all control surfaces locked in a neutral position? In other words, does the self-adjusting/ pendular pitch of the parafoil affect the IAS it stabilizes at?

Edit: Let's assume there is no control input from the pilot or any other changes taking place that are unrelated to altitude.

If the pitch trim is held constant, the IAS will be constant.
A rigid glider will control trim with an elevator trim tab.  The paraglider will control trim through slight shifts in the CG via tensioning the wires.
So long as the trim lift coefficient (and thereby angle of attack) is left fixed, the IAS will stay constant.
For a steady glide, we can write:
$L=W \cos(\theta)$
$-D=W \sin(\theta)$
We often think in terms of glide slope (instead of descent angle).
$GS=-\frac{1}{\tan(\theta)}$
Also,
$GS=\frac{L}{D}=\frac{C_L}{C_D}$
If we take our lift equation and put it in terms of lift coefficient:
$C_L=\frac{W \cos(\theta)}{q\,S_\mathrm{ref}}$
If trim is held constant, then $C_L$ is constant.
$W$ is constant (unless you drop ballast).
Since $C_L$ is constant, we're at constant $C_D$ (ignoring Reynolds number and Mach effects).  Which means we're at constant $L/D$.
Constant $L/D$ means constant descent angle $\theta$.
And $S_\mathrm{ref}$ the reference area of the wing is constant.
So, all that together means that we will glide at constant dynamic pressure - $q$.
$q=0.5 \rho V^2$
Where $V$ is TAS and $\rho$ is density.  Which shows that as you descend and $\rho$ increases, $V^2$ must decrease to maintain constant dynamic pressure.
But this is the exact relationship that is used to define IAS and EAS.  Constant EAS is defined as constant $q$.
IAS is EAS as measured by a particular installation -- but nothing about the installation is changing (most importantly angle of attack).  So, flight at constant $q$ is flight at constant EAS is flight at constant IAS.

Thanks! I'm a little unclear about the "pitch trim" you discuss. Are you saying your answer is valid only if the paraglider pilot takes conscious action to keep the trim constant? Or that a parafoil with a dead weight will automatically keep the trim constant?

Just to clarify, I'm asking about the case where there is no input from the pilot. I'm curious if the pitch of the parafoil changes at higher altitudes due to the reduced drag on the wing having more effect than the reduced drag on the smaller pilot.

@Oleg-- re last comment above-- the Drag force on any given part of the aircraft isn't reduced just because the TAS is higher for a given IAS.  L/D is constant, L = weight * cos glide angle, glide angle is determined by L/D.so D is constant.  If you want to incorporate that specific comment into question, I'll edit my answer to address.

@Oleg -- see my "Addendum" at end of my answer.

The pilot does not need to take conscious action to keep the trim constant.  If they hold the lines in the same place, they will hold it constant.
This may not be easy for a long period of time (if holding trim requires holding constant pressure) -- but if trim can be adjusted for 'hands off' flying, then it is the easy / default way to go.

Would the answer be different for a fixed wing glider that has all
control surfaces locked in a neutral position? In other words, does
the self-adjusting/ pendular pitch of the parafoil affect the IAS it
stabilizes at?

You seem to have an intuition that a parafoil (or paraglider or hang glider) tends to trim to a constant pitch attitude, and that this is somehow different from a more conventional aircraft's tendency to trim to a constant angle-of-attack.  In reality -- at least as long as we make no changes in the aircraft's configuration (physical shape), and we don't introduce Thrust into the picture, these are just two different ways of describing the same thing.
In an unpowered glider, the glide ratio with respect to the airmass is determined by the L/D ratio.  We can think of the glider's pitch attitude as being the result of the angle-of-attack minus the glide angle relative to the airmass; this result may be either positive or negative.  Also note that as long as we make no changes in the configuration (physical shape) of the aircraft, and as long as we don't introduce a thrust force from a motor, then each possible value of the L/D ratio, and thus each possible value of the glide angle relative to the airmass, corresponds to one particular angle-of-attack.  Therefore, tending to trim to a constant pitch attitude is the same as tending to trim to a constant angle-of-attack--either way, the glider will trim to a given glide ratio, and a given Indicated airspeed.  So the answer to your question is "no".
Now, if we were varying the glide ratio by deploying wing-mounted spoilers, or by adding some power, then in some hypothetical aircraft (perhaps with an autopilot in the loop?) that always trimmed to a given pitch attitude, that would generate a change in angle-of-attack, and therefore a large change in Indicated airspeed1.
Note also that if the spoilers (or the thrust force) generated no pitching moment relative to the aerodynamic center of the aircraft, they would generate a strong pitch torque relative to the CG of the aircraft, if the CG is far below the aerodynamic center of the aircraft.  This would result in a change in the trim angle-of-attack.
In the hang glider case specifically2, one valid way to analyze this situation is to view the wing itself as a free body, with a tendency to trim to given specific angle-of-attack, while the pilot's body exerts a pitch torque on the wing that (at least in steady-state flight) varies with the pitch attitude of the aircraft and thus tends to bias the angle-of-attack in a way that is dependent on pitch attitude.
Another valid way to analyze this situation is to simply note that the spoiler (or the thrust force) is generating a net pitching torque about the CG of the whole system, even if it is acting with zero moment-arm relative to the aerodynamic center of the wing.
If we introduce a thrust force that acts through the CG of the whole system, rather than through the aerodynamic center of system, we'll find that the wing still tends to trim to a constant angle-of-attack (and thus a nearly constant airspeed3) as we vary the thrust force, even though the pitch attitude will vary with the climb or descent angle.
So even though the "pendulum stability" provided by the arrangement of the CG far below the aerodynamic center of the wing does play an important role in the flight dynamics of parafoils, paragliders, hang gliders (in the case where pilot is holding a fixed position in the control frame), etc, it's not really true that such aircraft have an absolute tendency to trim to a fixed pitch attitude.  But for the simple case of steady-state4 wings-level gliding flight in a fixed configuration, that is a valid way to look at the dynamics.  We'll come up with a "no" answer to the questions quoted above regardless of whether we view the aircraft as tending to trim to a constant angle-of-attack, or to a constant pitch attitude.
Addendum: a comment by the questioner has indicated that he believes that Drag may be reduced during flight at high altitudes, and wonders if the Drag is reduced proportionally on all parts of the aircraft, and if not, whether this would affect the aircraft's trim angle-of-attack, considering the large vertical distance between the various parts of the aircraft.  In steady-state gliding flight, L = W * cosine (glide angle) = W * cosine (arctan (D/L)), and D = W * (sine (glide angle)) = W * sine (arctan (D/L)).  Therefore in steady-state gliding flight, the L and D values are completely determined by the L/D ratio which is completely determined by the angle-of-attack, regardless of how high the Indicated Airspeed must be to generate those values of L and D.5  Therefore there is no reduction in drag on any part of the aircraft just because the altitude is increased.
Read more about "pendulum" pitch stability in this answer to a related ASE question: Does "pendulum effect" apply to hang gliders or any aircraft?
Footnotes:

Consider an aircraft whose trim characteristics are such that it perfectly holds a given angle-of-attack regardless of changes in thrust, or changes in configuration that increase the drag coefficient (without affecting the lift coefficient).  Even such an aircraft will still experience some change in Indicated airspeed when the Thrust is altered, or when the drag coefficient is altered.  The reason for this is that the total Lift required for steady-speed (wings-level) flight is not constant, but rather varies according to the cosine of the descent or climb angle.  For more, see this related ASE answer: Does lift equal weight in a climb?

We refer to a hang glider specifically, because the wing shape of a paraglider or parafoil may have no tendency to trim to any given angle-of-attack without the pilot attached, even the wing could somehow be made to maintain its shape.  Note also that we're assuming the pilot either has his "hang strap" attached somewhat below the CG of the wing itself, or we're assuming that the pilot is holding himself in a fixed position relative to the control frame.  If neither of these things are true-- if the pilot's weight is attached exactly at the CG of the wing, and the pilot is just hanging freely  and exerting no muscle force on the control frame, then his weight effectively acts at the CG of the wing, not below it.  In a paraglider or parafoil, the multiple lines act as fixed struts that hold the pilot in a completely fixed position relative to the wing, unless he is making intentional control inputs by pulling on the lines etc, so in way these aircraft provide an even better example of the dynamics we're exploring, except for the lack of inherent pitch stability in the wing itself -- these aircraft are totally dependent on "pendulum stability*.

See footnote (1).

Re "steady-state": for further insight into what "pendulum stability" does and does not mean, search up some videos of hang glider and paraglider aerobatics.  Though limited to positive G-loadings, the maneuvers that can be performed in such aircraft may surprise the reader!

Ignoring any variation in lift and drag coefficients due to Reynolds number.  Note that if the drag coefficients of the pilot and the wing are affected differently by the change in Reynolds number due to decreased air density and increased TAS, then that could produce a change in the trim angle-of-attack and L/D ratio, even if the L/D ratio corresponding to any given angle-of-attack remained unchanged.  Note also that if we're bringing Reynold's number effects into the picture, we can't assume that the L/D ratio corresponding to any given angle-of-attack will remain unchanged.  So-- "it's complicated".  But the effect of changes in Reynold's number on the lift and drag coefficients appears to be beyond the intended scope of the  question.

Thank you! Two great explanations. I wish I could accept both, since I'm using both.

@Oleg -- you are welcome, no problem, I understand you need to choose 1 --

@Oleg -- I haven't thoroughly thought through yet the case where, if Drag were somehow reduced, and were reduced proportionally on all parts of aircraft, and Lift were reduced proportionally as well (so no change in L/D for given angle-of-attack), that might affect trimmed angle-of-attack.  Seems it might be equivalent to  change in pilot weight-- i.e. the relevant balance between aerodynamic torques and torques due to pilot weight below CG might be changed.  But, I think it's not relevant to actual question, because in reality (at least if we ignore Reynolds number effects) the drag (ctd)

(ctd) the Drag forces are not changed.


text:
What determines the best glide speed?
I found the best glide speed of cessna 152 is 60 kts. My question is why it is not more or less than 60 kts? What are the factors upon which the best gliding speed is determined?

Because that gives the most distance per unit drop. A lower speed will have a steeper descent and a higher speed will also have a steeper descent. Remember when you are gliding, your only speed control is pitch, and pitch also affects your descent rate. Its also not that simple as you state, best glide is dependent on the weight of the aircraft, usually the POH states "best glide" at maximum weight.

I understand a higher speed will steeper may descent more than a lower speed. But how that controls the best gliding speed? Does it only depend on the weight? If my weight is less/more how that will control the best gliding speed?

do these posts answer your question? https://aviation.stackexchange.com/q/606/1467 https://aviation.stackexchange.com/q/3610/1467

The weight doesn't "control" the best glide speed, it dictates it. Best glide speed means the most distance traveled per distance dropped. The lower the weight, the lower the speed. See this FAA document for some more information.

Basically, the faster you go, the more lift and drag a wing will produce. These two values are not proportional, though. When speeding up, the amount of drag produced is higher than the extra amount of lift - which is why you need additional power to maintain level at higher speeds.
When you go slower, the amount of drag will reduce more than the amount of lift produced - at least for a while. That's why going slow is better in terms of gliding distance. The amount of "drag per lift" is very low. However, slowing down beyond a certain point, the wing will rapidly start producing less lift, because airflow separates from the wing. This is what's known as a stall. The best glide speed is the speed as which the drag is as low as possible while the wing is still producing a relatively large amount of lift.
This is illustrated on a speed polar, like this one:

The black line indicates the rate of sink for a given airspeed. The optimum glide speed is the speed corresponding to the point where the red line touches the black line (Vbg).
The red line is a straight line going from (0,0) and touching the speed polar at exactly one point.
A change in aircraft mass will shift the curve along the vertical axis, which is why a heavier aircraft has a higher best glide speed than a lighter one. The intersecting point between the black and red line would shift to the right as the black line is shifted down, and vice versa.

Now I have one question,  according to graph at Vmd the sink rate is less and by this I can remain up in the air for more time. And at Vbg sink rate is greater. Then why my best glide speed is not at the speed of Vmd? Why it is a higher speed at which sink rate is more?

At Vmd you are indeed sinking slower, but you are also moving forward slower. Your air time will be higher, but you will cover a shorter distance because your forward speed is slower. Vmd/Minimum sink is the speed that will keep you in the air for the longest time. The Vbg/best glide is the speed that will let you cover the greatest distance.

This is kind of a fine point, but in general, slowing down does not actually cause a reduction in Lift.  See my recent answer for a different view.

@quiet flyer that’s very much a valid point - unless accelerating upwards or downwards the lift must equal the weight of the aircraft.  However, it does depend on which variables you change and which you don’t - if the airspeed is reduced and the attitude is unchanged then you’re right, but if the angle of incidence is maintained then the aircraft will accelerate downwards (initially at least).

The most important factors for the best glide speed are the aircraft's wing loading, the air density, the wing's aspect ratio, and the aerodynamic quality of the aircraft.
The aircraft must create lift equal to its own weight. The drag for doing so varies with airspeed, and to find the point where the glide ratio has its maximum, drag must be minimal. To find this speed, we describe drag mathematically as the sum of two components:

Parasitic drag, which goes up with the square of airspeed. We express this as the zero-lift drag, a drag component which is independent of lift: $D_0 = \frac{\rho}{2}\cdot v^2\cdot S\cdot c_{D0}$
Lift-dependent or induced drag which goes down with the inverse of the square of airspeed: $D_i = \frac{\rho}{2}\cdot v^2\cdot S\cdot \frac{c_L^2}{\pi\cdot AR\cdot\epsilon}$

Now it helps to find the lift coefficient to create the needed lift at a given speed:
$$c_L = \frac{m\cdot g}{\frac{\rho}{2}\cdot v^2\cdot S}$$
Which, when inserted into the formula for induced drag, produces
$$D_i = \frac{(m\cdot g)^2}{\frac{\rho}{2}\cdot v^2\cdot S\cdot \pi\cdot AR\cdot\epsilon}$$
Now it should be obvious that induced drag is indeed proportional to the inverse of flight speed squared. We can simplify this a little by inserting $AR = \frac{b^2}{S}$ and express the total drag as the sum of both components:
$$D = \frac{\rho}{2}\cdot v^2\cdot S\cdot c_{D0} + \frac{(m\cdot g)^2}{\frac{\rho}{2}\cdot v^2\cdot \pi\cdot b^2\cdot\epsilon}$$
Next, we differentiate with respect to speed $v$ and need to set the result to zero to arrive at an equation for the speed of lowest drag:
$$\frac{∂ D}{∂ v} = \rho\cdot v\cdot S\cdot c_{D0} - \frac{(2\cdot m\cdot g)^2}{\rho\cdot v^3\cdot \pi\cdot b^2\cdot\epsilon} = 0$$
$$\rho\cdot v^4\cdot S\cdot c_{D0} = \frac{(2\cdot m\cdot g)^2}{\rho\cdot \pi\cdot b^2\cdot\epsilon}$$
$$v = \sqrt[4]{\frac{(2\cdot m\cdot g)^2}{\rho^2\cdot\pi\cdot b^2\cdot\epsilon\cdot S\cdot c_{D0}}}$$
$$v = \sqrt{\frac{2\cdot m\cdot g}{\rho\cdot S\cdot\sqrt{\pi\cdot AR\cdot\epsilon\cdot c_{D0}}}}$$
There you have it: The best glide speed is proportional to the square root of both the wing loading $\frac{m\cdot g}{S}$ and the inverse of air density $\rho$, and the fourth root of the inverse of the aspect ratio $AR$, the Oswald factor $\epsilon$ and the zero-lift drag coefficient $c_{D0}$. The Oswald factor is a measure of the quality of lift production and is close to unity in most cases.
Thankfully, quiet flyer reminds me in the comments that I need to point out that I always talk about true air speed here. If you want to set your speed using the speed indicator, reduce the value by the applicable density correction.
Nomenclature:
$c_{D0} \:$  zero-lift drag coefficient
$c_L \:\:\:$ lift coefficient
$S \:\:\:\:\:$  reference area (wing area in most cases)
$v \:\:\:\:\:$  airspeed
$\rho \:\:\:\:\:$  air density
$\pi \:\:\:\:\:$  3.14159$\dots$
$AR \:\:$  aspect ratio of the wing
$\epsilon \:\:\:\:\:$  the wing's Oswald factor
$m \:\:\:\:$  the aircraft's mass
$g \:\:\:\:\:$  gravitational acceleration
$b \:\:\:\:\:$  wingspan

Is this the same as the L/D max speed (Vldmax)?

@MaxvonHippel: Yes. Minimum drag at constant lift means that L/D is at its maximum.

Re "There you have it: The best glide speed is proportional to the square root of both the wing loading m⋅gS and the inverse of air density ρ..."  -- since you have brought air density into it, it might be helpful to specify that you are speaking of TAS (if you are).

@quietflyer Yes, I am. Good point!

(it's simpler than it may looks)
If you are at a certain height, you have a certain amount of potential energy (or height energy). The only thing you can do is to convert it to kinetic energy (or speed, which then creates lift). The problem: drag takes up energy too. So all the energy you loose due to drag means a loss in kinetic energy (=speed) and therefore a loss in lift.
The question actually is: how to reduce drag to a minimum?
It's actually quite simple:
there are roughly two different kinds of drag:

induced drag, induced by the angle of attack of the airplane. The more your nose goes up (so the lower your airspeed is), the higher the induced drag. This is an exponential relation.

parasitic drag, comes from the air and is the "usual" drag you also feel with a car or bike. It depends exponential on the airspeed.

The total drag consists of the sum of both.
The minimum is the best gliding speed.

Wouldn't the best glide speed be a bit faster than the minimum drag speed (since, by definition, the airplane covers more distance per unit time at higher speeds?)

Sure. But your goal is not to fly the longest distance in the shortest time, which means speed is irrelevant, only the efficiency matters. If you loose, say, 500 feet, you better need 2 minutes for that with a speed of 50 knots instead of 1 minute with a speed of 70. We only look for the best height-loss versus distance covered ratio. We do not care about the time at all, it is completely irrelevant.

@reirab --no.  We can show that the best glide angle occurs at the highest ratio of L to D.  We can also show that this occurs exactly at the point of minimum D.  (By the way, L is nearly constant and nearly equal to Weight; more precisely, L = Weight * cosine glide angle.)

This is kind of a fine point, but in general, slowing down does not actually cause a reduction in Lift.

Also, if we were really trying to minimize the rate of energy loss, I think we'd end up at the airspeed for min sink rate, not max glide ratio. So, generally good answer but some room for improvement--

@quietflyer, I read your comments but I actually miss somehow what can be improved in the answer. What exactly is the point that you are trying to make? I think the answer states clearly that it is not about the speed but about the efficiency, right?

I've never heard of the term maximum gliding speed, there's no special limitation to how fast you can fly a c152 without an engine as opposed to with it working. I think what you are talking about is best glide speed, also known as Vbg, which is the speed that gives you the farthest horizontal distance travelled per unit of height lost. If I remember correctly 60kts is the best glide with flaps extended, 65kts was best glide with no flaps.
The best glide speed actually varies based on weight, as do most of the V-speeds. A heavier airplane would mean a faster Vbg and a lighter one a slower Vbg. On a c152 the difference is pretty small, maybe 2 kt either way, so giving a 1 speed answer makes sense as it's easy to remember. Best glide speed on a big airplane will vary much more and would need to be calculated based on weight estimate at that point in the flight.

In this answer, we'll confine ourselves to steady linear flight with the wings level.
We can show that the highest glide ratio (i.e. smallest or flattest glide angle) occurs at the highest ratio of L (Lift) to D (Drag).  In fact we can show that the glide ratio is exactly the same as the ratio of L to D.  We can also show that the highest ratio of L to D occurs exactly at the point of minimum D. (L is nearly constant and nearly equal to Weight; more precisely, L = Weight * cosine glide angle.)

I found the best glide speed of cessna 152 is 60 kts. My question is why it is not more or less than 60 kts? What are the factors upon which the best gliding speed is determined?

Because that is where D (Drag) is at a minimum.  Nothing else matters.
Keep in mind that Drag is proportional to the product of the drag coefficient-- which increases as we increase the angle-of-attack-- and the square of the airspeed.
Other answers give additional insight as to why the point of minimum Drag occurs where it does-- it is the point where Induced Drag and Parasite Drag are exactly equal.
By the way, the ratio of L/D is exactly equal to the ratio of Cl / Cd, where Cl and Cd are the coefficients of Lift and Drag.  Unlike Lift, Cl is not nearly constant across a wide range of angles-of-attack or airspeeds.  It is always much  greater at high angles-of-attack (low airspeeds), at least until we get very close to the stall angle-of-attack.  Cd is also always much greater at a high angle-of-attack (low airspeed) than at a low angle-of-attack (high airspeed). So we can also say that the  maximum glide ratio occurs at the maximum ratio of Cl to Cd.  Lest there be any confusion, we should also observe that this is the point of minimum Drag, but not the point of minimum drag coefficient.  If we were flying at the angle-of-attack for the maximum glide ratio, and we then moved the stick or yoke forward to reduce the angle-of-attack (and increase the airspeed), this would reduce the drag coefficient (as well as the lift coefficient), but this would also increase Drag, and would decrease L/D and Cl / Cd.
In comments you indicate that you want to know how best glide speed varies with weight.  A good approximation is that it scales up or down in proportion to the square root of any change in your wing loading.  In gliders where we don't have drag from a windmilling prop to complicate things-- or in any airplane where we've stopped the prop-- this should be exactly true, except for effects due to changes in Reynolds number.
For completeness, we should say something about where the minimum power-off sink rate occurs. That turns out to happen at the angle-of-attack that gives the maximum value of (Cl cubed) / (Cd squared).  This always occurs at a higher angle-of-attack (lower airspeed) than does the best glide ratio.

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